For some `x in [-1, 1]` we have `cos^(-1)x =tan^(-1)x`.
The value of `sin(cos^(-1)x)` is:

Options

A. `(sqrt(5)-1)/(2)`
B. `(sqrt(5)+1)/(2)`
C. `(sqrt(2)-1)/(2)`
D. `(sqrt(2)+1)/(2)`

Answer

To solve the problem, we start with the equation given in the question:

\[ \cos^{-1}(x) = \tan^{-1}(x) \]

### Step 1: Express \(\tan^{-1}(x)\) in terms of \(\cos^{-1}(x)\)

From the identity of inverse trigonometric functions, we know that:

\[ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \]

If we let \(\theta = \cos^{-1}(x)\), then:

\[ \tan(\theta) = \frac{\sin(\cos^{-1}(x))}{x} \]

### Step 2: Use the Pythagorean identity

Using the Pythagorean identity, we can express \(\sin(\cos^{-1}(x))\):

\[ \sin(\cos^{-1}(x)) = \sqrt{1 - x^2} \]

### Step 3: Set up the equation

Since \(\tan^{-1}(x) = \theta\), we can write:

\[ x = \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{\sqrt{1 - x^2}}{x} \]

### Step 4: Cross-multiply and simplify

Cross-multiplying gives:

\[ x^2 = \sqrt{1 - x^2} \]

Squaring both sides results in:

\[ x^4 = 1 - x^2 \]

### Step 5: Rearranging the equation

Rearranging the equation gives us:

\[ x^4 + x^2 - 1 = 0 \]

### Step 6: Let \(y = x^2\)

Substituting \(y = x^2\), we get a quadratic equation:

\[ y^2 + y - 1 = 0 \]

### Step 7: Solve the quadratic equation

Using the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

\[ y = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{5}}{2} \]

### Step 8: Find the positive root

Since \(y = x^2\) must be non-negative, we take the positive root:

\[ y = \frac{-1 + \sqrt{5}}{2} \]

### Step 9: Substitute back to find \(x^2\)

Thus, we have:

\[ x^2 = \frac{-1 + \sqrt{5}}{2} \]

### Step 10: Find \(\sin(\cos^{-1}(x))\)

Now we can find \(\sin(\cos^{-1}(x))\):

\[ \sin(\cos^{-1}(x)) = \sqrt{1 - x^2} = \sqrt{1 - \frac{-1 + \sqrt{5}}{2}} = \sqrt{\frac{3 - \sqrt{5}}{2}} \]

### Final Answer

The value of \(\sin(\cos^{-1}(x))\) is:

\[ \sin(\cos^{-1}(x)) = \sqrt{\frac{3 - \sqrt{5}}{2}} \]

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